A tuning fork, vibrating with a frequency of 90 Hz. is moving towards an observer with a with a speed of 0.1 times the speed of sound. The frequency of the note heard by the observer would be:

Correct option is C

​$$\textbf{Given:} \\\text{The source (tuning fork) is moving \textbf{towards the observer}.} \\\bullet \text{ Source frequency, } f_0 = 90\, \text{Hz} \\\bullet \text{ Source speed, } V_s = 0.1 \times V_{\text{sound}}\\[10pt]\text{When the \textbf{source moves towards} a stationary observer, the apparent frequency is given by:} \\f = f_0 \left(1 + \frac{V_s}{V_{\text{sound}}}\right)\\[10pt]f = 90 (1 + 0.1) \\f = 90 \times 1.1 = 99\, \text{Hz} \approx 100\, \text{Hz}$$​​